A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB

Answer:

1)copper wire

Explanation:

it is the best electric conductor

Answer:

0.00000009512

Explanation:

Scientific notation is a very useful and abbreviated way of writing quantities that are very large or small. It consists of placing the number with an integer and multiplying by an exponent to arrive at the same number.

let's pass the number 9,512 10⁻⁸ to decimal notation

       9,512 / 10⁸ = 9,512 / 100000000

        0.00000009512

As we see writing this number, it is very easy to make mistakes

Answer:

i) A dimension is the physical nature of a quantity. The two limitations of dimensional analysis is as following:

ii) Given:

Velocity = v

Time = t

Force = F

Force = mass x acceleration

         = mass x velocity/time

So, mass= (force x time) / velocity

[mass] = Ftv^-1

Hence, dimesnion of mass is Ftv^-1.

Answer:

Salt

Explanation:

Salt plays a crucial role in maintaining human health. It is the main source of sodium and chloride ions in the human diet. Sodium is essential for the nerve and muscle function and is involved in the regulation of fluids in the body. Sodium also plays a role in the body's control of blood pressure and volume. Salt is harvested by seawater or brine is fed into large ponds of water and is drawn out through natural evaporation which allows the salt to be subsequently harvested.

Have a good day and stay safe!

Answer:

Explanation:

  P = VI

  I = P/V = P/120

  R = V/I = V/(P/V) = V^2/P = 14400/P

Toaster: I = 1800/120 = 15 . . . amps

  R = 14400/1800 = 8 . . . ohms

Fry pan: I = 1300/120 = 10.833 . . . amps

  R = 14400/1300 = 11.08 . . . ohms

Lamp: I = 100/120 = 0.833 . . . amps

  R = 14400/100 = 144 . . . ohms

The total current exceeds 20 A, so will blow the fuse.

Answer:

Explanation:

Distance between fringe or fringe width =  xλ /  d

where x is location of screen and d is slit separation

Given x = 4 m

λ = 694 nm

d = .085 x 10⁻³ m

distance between fringes

= 4 x 694 x 10⁻⁹ / .085 x 10⁻³

= 4 x 694 x 10⁻⁹ / 85 x 10⁻⁶

= 32.66 x 10⁻³ m

= 32.66 mm .

3.267 cm

b )

when submerged in water , wavelength in water becomes as follows

wavelength in water = wave length / refractive index

= 694 / 1.333 nm

= 520.63 nm

new distance between fringes

3.267 / 1.333

= 2.45 cm .

Answer:

the angular speed of the person increases, being able to make more turns and faster.

 K₂ = K₁ I₁ / I₂

Explanation:

When the divers are turning the system is isolated, so all the forces are internal and therefore also the torque, therefore the angular momentum is conserved

initial, joint when starting to turn

         L₀ = I₁ w₁

final. When you shrink your arms and legs

         Lf = I₂ w₂

         L₀ = Lf

         I₁ w₁ = I₂ w₂

when you shrink your arms and legs the distance to the turning point decreases and since the moment of inertia depends on the distance squared, the moment of inertia also decreases

      I₂ <I₁

         w₂ = I₁ / I₂ w₁

therefore the angular speed of the person increases, being able to make more turns and faster.

When it goes into the water it straightens the arm and leg, so the moment of inertia increases

          I₁> I₂

           w₁ = I₂ / I₁ w₂

therefore we see that the angular velocity decreases, therefore the person trains the water like a stone and can go deeper faster.

In both cases the kinetic energy is

          K = ½ I w²

the initial kinetic energy is

          K₁ = ½ I₁ w₁²

the final kinetic energy is

          K₂ = ½ I₂ w₂²

we substitute

          K₂ = ½ I₂ (I₁ / I₂ w1² 2

          K₂ = ½ I₁² / I₂ w₁² = (½ I₁ w₁²)  I₁ / I₂  

          K₂ = K₁ I₁ / I₂

therefore we see that the kinetic energy increases by factor I₁/I₂

Answer:

The time taken is  [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]

Explanation:

From the question we are told that

     The speed of the current is  [tex]v__{R}}[/tex]

     The speed of the swimmer in direction of current is [tex]v__{S}}[/tex]

      The distance traveled by the swimmer is  [tex]D[/tex]

       The time taken to travel this distance is  [tex]t_{out}[/tex]

      The speed of the swimmer against  direction of current is  [tex]v__{s}}[/tex]

The resultant speed for downstream current is

       [tex]V_{r} = v__{S}} +v__{R}}[/tex]

The time taken can be mathematically represented as

      [tex]t_{out} = \frac{D}{V_{r}}[/tex]

      [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]

Answer:

F = 423.63 N

Explanation:

Since, the x-component and y-components of the acceleration of ball are given. Therefore, we need to find the resultant or net acceleration of the soccer ball first. For that purpose we use to the formula for the resultant of rectangular components of a vector:

a = √(ax² + ay²)

where,

a = net acceleration = ?

ax = x - component of acceleration = 950 m/s²

ay = y - component of acceleration = 750 m/s²

Therefore,

a = √[(950 m/s²)² + (750 m/s²)²]

a = 1210.4 m/s²

Now, from Newton's Second Law, we know that:

F = ma

where,

m = mass of ball = 0.35 kg

F = Net force acting on ball = ?

F = (0.35 kg)(1210.4 m/s²)

F = 423.63 N

Answer:

100 J

Explanation:

Work = change in energy

W = ΔKE

W = ½ mv² − ½ mv₀²

W = ½ m (v² − v₀²)

W = ½ (10 kg) ((6 m/s)² − (4 m/s)²)

W = 100 J

Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

Given data:

[tex]t_A=0.55s[/tex]

[tex]t_B=0.45s[/tex]

(a) The distance from the kicker to each of the 2 spectators is given by:

[tex]d_A=v \times t_A[/tex]

where,

v= speed of sound

[tex]t_A[/tex]=time taken for the sound waves to reach the ears

[tex]d_A=343\times 0.55=188.65[/tex]m

(b)[tex]d_B=v \times t_B[/tex]

where,

v= speed of sound

[tex]t_B[/tex]=time taken for the sound waves to reach the ears

[tex]d_B=343\times 0.45=154.35m[/tex]

(c)As the angle b/w slight lines  from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

[tex]D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m[/tex]

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it =  λ dx where  λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x  / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

=  k λ dx  .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫  k λ  .15  / (x² + .15² )³/² dx

=  k λ  x L / .15 √( L² / 4 + .15² )

b) The length of the rod:

[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )[/tex]

Given:

d = 1.5 mλ = 2.5 nC/m

Let the plastic rod extends from - L to + L .Consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .It will create a field at point P on y -axis.

Distance of point P =[tex]\sqrt{x^2 + 0.15^2}[/tex]

E.F at P due to small charged length[tex]dE = \frac{ k \lambda x.dx}{(x^2 + .15^2 )}[/tex]

Its component along Y - axis = dE cosθ where θ is angle between direction of field dE and y axis

[tex]= \frac{dE x .15 }{\sqrt{x^2 + .15^2} }\\\\= \frac{k \lambda dx .15}{(x^2 + .15^2 )^{1/2}}[/tex]

If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . We can say that the component of field in perpendicular to y axis will cancel out each other.

Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )}[/tex]

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Answer:

20 Yards

Explanation:

|---20----|

|            |

| 50       |50

|---D--->|

Start      End

Total displacement(D)  20 yards (East).

Explanation:

2.  No, not always.  Normal force is equal to force of gravity only when there's no acceleration in the vertical direction.

For example, when you stand in an elevator that's not moving, or moving at constant speed, then the normal force equals your weight.  But when the elevator accelerates upward, the normal force increases (making you feel heavier).  And when the elevator slows down, the normal force decreases (making you feel lighter).

3.  Yes, it is possible for an object to be moving eastward and experience a net force westward.  An example is a car applying the brakes.

4.  Friction force allows you to walk.  When you push against the floor, the floor's friction pushes back, as Newton's third law says.

If you try to walk on a slippery surface like ice, you won't be able to push against the ice, and the ice won't push back.

Answer:

(a) 498.6 Hz

(b) 534.6 Hz

Explanation: Please see the attachments below

Answer:

d = 41.91 m

Explanation:

In order to calculate the distance traveled by the train while superman passes it, you write the equations of motion for both superman and train:

For train, you have a motion with constant speed. You write the equation of motion of the position of the front of the train:

[tex]x=x_o+v_1t[/tex]    (1)

xo: initial position of the front of the train = 500m

v1: speed of the train = 50km/h

For superman, you take into account that the motion is an accelerated motion (you assume superman is at the origin of coordinates):

[tex]x'=v_2t+\frac{1}{2}at^2[/tex]   (1)

v2: initial speed of superman = 100km/h

a: acceleration = 10m/s^2

When superman passes the train, both positions x and x' will be equal. Hence, you equal the equations (1) and (2) and you calculate the time t. But before you convert the units of the velocities v1 and v2 to m/s:

[tex]v_1=50\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=13.88\frac{m}{s}\\\\v_2=100\frac{km}{h}=\frac{1000m}{1km}*\frac{1h}{3600s}=27.77\frac{m}{s}[/tex]

Thus, you equal x=x'

[tex]x=x'\\\\x_o+v_1t=v_2t+\frac{1}{2}at^2\\\\500m+(13.88m/s)t=(27.77m/s)t+\frac{1}{2}(10m/s^2)t^2\\\\(50\frac{m}{s^2})t^2+(13.89\frac{m}{s})t-500m=0[/tex]

You solve the last equation for t by using the quadratic formula:

[tex]t_{1,2}=\frac{-13.89\pm \sqrt{(13.89)^2-4(50)(-500)}}{2(50)}\\\\t_{1,2}=\frac{-13.89\pm 316.53}{100}\\\\t_1=3.02s\\\\t_2=-3.30s[/tex]

You only use t1 = 3.02s because negative times do not have physical meaning.

Next, you replace this value of t in the equation (1) to calculate the position of the train (for when superman just passed it):

[tex]x=500m+(13.88m/s)(3.02s)=541.91m[/tex]

x is the position of the front of the train, then, the dstance traveled by the train is:

d = 541.91m - 500m = 41.91 m

Answer:

Explanation:

If there is only 1 force, the body can never be in equilibrium, providing that the force is not zero (and that would hardly be a force. Zero is possible in math and it means something. It is debatable in physics).

You cannot think of a condition where something is stationary on planet earth and there are not 2 forces or more forces involved.

Think of something like a block of wood sitting on a table. It is not moving, we'll say. Gravity is holding it down, but what is pushing up on it?

The table is. There are 2 forces and they are equal in magnitude, but opposite in direction. That matters.

Answer:

So if its acceleration is -2m/s^2 that means every second the initial velocity would be subtracted by 2. So since it took 3 seconds 2*3=6. The initial velocity was 6 m/s

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m  (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

The value of Ey such that the particle will cross the x axis at x=1.5 cm is -4556.18 N/m.

The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.

Given is a particle leaves the origin with a speed of 3.6 x 10⁶ m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis.

The distance x = 1.5 cm = 1.5×10⁻² m (assumed, not given in question)

The horizontal distance traveled by particle is

x = ucosθt

t = x/ucosθ

The force in an electric field is F = qE...................(1)

where, q is charge , E is the strength of electric field

From, newton 2nd law of motion, Force F = ma.................(2)

Equating both the equations, we get

ma = qE

a = qE/m..................(3)

The vertical distance, y =usinθt - 1/2at²

From equation 3, we have

y = usinθt  -  1/2 (qE/m) t²

if y = 0, t = 2musinθ/(qE) = x / (ucosθ)

The electric field is represented as

Also, E = 2mu²×sinθ×cosθ/(xq)

Plug the values, we get

E = 2×(9.1×10⁻³¹)×(3.6 x 10⁶)²×sin34°×cos34°/( 1.5×10⁻² ×(-1.6)×10⁻¹⁹)

E = -4556.18 N/m

Thus, the electric field of the particle is  -4556.18 N/m.

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Answer:

The value of F= - 830 N

Since the force is negative, it implies direction of the force applied was due south.

Explanation:

Given data:

Mass = 1000-kg

Distance, d = 240 m

Initial velocity, v1 = 20.0 m/s

Final velocity, v2 = 0 (since the car came to rest after brake was applied)

v2²= v1² + 2ad (using one of the equation of motion)

0=  20² + (2 x a x  240)

0= 400 + 480 a

a = - 400/480

a = - 0.83 m/s²

Then, imputing the value of a into

F = ma

F = 1000 kg x ( - 0.83 m/s²)

F= - 830 N

The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.

Answer:

The  normal force N1 exerted by the floor is  [tex]N_1 = 951 \ N[/tex]

The  normal force N2 exerted by the wall is  [tex]N_2= 616.43 \ N[/tex]

The frictional force exerted by the wall is  [tex]f = N_2 = 616.43 \ N[/tex]  

Explanation:

From the question we are told that

    The length of the ladder is  [tex]L = 4.6 \ m[/tex]

    The weight of the ladder  is

    The distance of the ladder position on the wall from the floor is  [tex]D = 3.75 \ m[/tex]

     The mass of the person is  [tex]m = 90 kg[/tex]

Applying Pythagoras theorem

The length of the position the ladder on the ground from the base of the wall is

    [tex]A = \sqrt{L^ 2 - D^2}[/tex]

substituting values

    [tex]A = \sqrt{(4.6^2)-(3.75^2)}[/tex]

    [tex]A = 2.66 \ m[/tex]

  In order the for the ladder not to shift from the ground the sum of the moment about the position of the ladder on the ground must be equal to zero this is mathematically represented as

        [tex]\sum M = 0 = N_2 * D - [\frac{1}{2} * W_L ] * [(mg) *A ][/tex]

         [tex]\sum M = 0 = N_2 * 3.75 - [\frac{1}{2} * 69.0 ] * [(90*9.8) * \frac{4.6}{2.66} ][/tex]

        [tex]N_2 * 3.75 =2311.62[/tex]

        [tex]N_2 * 3.75 =2311.62[/tex]

        [tex]N_2= 616.43 \ N[/tex]

Now the force exerted by the floor on the ladder is mathematically represented as

           [tex]N_1 = W_L + (m * g )[/tex]

substituting values

          [tex]N_1 = 951 \ N[/tex]

Now the horizontal forces acting on the ladder are [tex]N_2 \ and \ f[/tex] and they are in opposite direction so

     [tex]f = N_2 = 616.43 \ N[/tex]  

Explanation:

It is given that,

Total mass is 70 kg

The truck exerts a constant force of 20 N.

Then the net force is given by :

F = ma

a is acceleration of rider

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2[/tex]

Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :

[tex]v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s[/tex]

Since, 1 m/s = 2.23 mph

4.28 m/s = 9.57 mph

So, the speed of the rider is 4.28 m/s or 9.57 mph.  

Answer:

15N

Explanation:

F=ma

m=1500g = 1.5kg

a=10m/s2

1.5×10=15 N

Answer:15000gms^-2

Explanation:

F=m×a

m=1500g, a=10ms^-2

F=(1500×10)gms^-2

F=15000gms^-2

Answer:

48 Nm

Explanation:

Moment, or torque, is the cross product of radius and force vectors.

τ = r × F

τ = (0.80 m) (60 N)

τ = 48 Nm

Answer:

Explanation:

During slowing down , initial angular velocity ω₁ = 22 rad /s

final angular velocity ω₂ = 13.5 rad /s

using the law's of motion formula for rotation

ω₂² =  ω₁² + 2 αθ  , α is angular acceleration and θ is angle in radian rotated during this period

13.5² = 22² - 2xα x 13.8

2xα x 13.8 = 484 - 182.25

α  =  10.93 rad / s²

Explanation:

A) Draw free body diagrams of both blocks.

Force P is pushing right on block A, which will cause it to move right along the incline.  Therefore, friction forces will oppose the motion and point to the left.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force N pushing up and left 10° from the vertical,

Friction force Nμ pushing down and left 10° from the horizontal,

Reaction force Fab pushing down,

and friction force Fab μ pushing left.

There are 2 forces acting on block B:

Reaction force Fab pushing up,

And elastic force kx pushing down.

(There are also horizontal forces on B, but I am ignoring them.)

Sum of forces on A in the x direction:

∑F = ma

P − N sin 10° − Nμ cos 10° − Fab μ = 0

Solve for N:

P − Fab μ = N sin 10° + Nμ cos 10°

P − Fab μ = N (sin 10° + μ cos 10°)

N = (P − Fab μ) / (sin 10° + μ cos 10°)

Sum of forces on A in the y direction:

N cos 10° − Nμ sin 10° − Fab = 0

Solve for N:

N cos 10° − Nμ sin 10° = Fab

N (cos 10° − μ sin 10°) = Fab

N = Fab / (cos 10° − μ sin 10°)

Set the expressions equal:

(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)

Cross multiply:

(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)

Distribute and solve for Fab:

P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)

P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)

Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)

Sum of forces on B in the y direction:

∑F = ma

Fab − kx = 0

kx = Fab

x = Fab / k

x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))

Plug in values and solve.

x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))

x = 0.0408 m

x = 4.08 cm

B) Draw free body diagrams of both blocks.

Force P is pushing block A to the right relative to the ground C, so friction force points to the left.

Block A moves right relative to block B, so friction force on A will point left.  Block B moves left relative to block A, so friction force on B will point right (opposite and equal).

Block B moves up relative to the wall D, so friction force on B will point down.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force Fc pushing up,

Friction force Fc μ₁ pushing left,

Reaction force Fab pushing down and left 15° from the vertical,

and friction force Fab μ₂ pushing up and left 15° from the horizontal.

There are 5 forces acting on block B:

Weight force 750 n pushing down,

Normal force Fd pushing left,

Friction force Fd μ₁ pushing down,

Reaction force Fab pushing up and right 15° from the vertical,

and friction force Fab μ₂ pushing down and right 15° from the horizontal.

Sum of forces on B in the x direction:

∑F = ma

Fab μ₂ cos 15° + Fab sin 10° − Fd = 0

Fd = Fab μ₂ cos 15° + Fab sin 15°

Sum of forces on B in the y direction:

∑F = ma

-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0

Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Substitute:

(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750

Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)

Sum of forces on A in the y direction:

∑F = ma

Fc + Fab μ₂ sin 15° − Fab cos 15° = 0

Fc = Fab cos 15° − Fab μ₂ sin 15°

Sum of forces on A in the x direction:

∑F = ma

P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0

P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁

Substitute:

P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁

P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°

P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)

First, find Fab using the given values.

Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)

Fab = 1151.9 N

Now, find P.

P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)

P = 1095.4 N

Answer:

N = 243.596 N ≈ 243.6 N

Explanation:

mass of person = 69 kg ( M )

mass of aluminium ladder = 11 kg ( m )

length of ladder = 6.4 m ( l )

base of ladder = 2 m from the house (d )

center of mass of ladder = 2 m from the bottom of ladder

person on ladder standing = 3 m from bottom of ladder

Calculate the magnitudes of the forces at the top and bottom of the ladder

The net torque on the ladder = o ( since it is at equilibrium )

assuming: the weight of the person( mg) acting at a distance x along the ladder. the weight of the ladder ( mg) acting halfway along the ladder and the reaction N acting on top of the ladder

X = l/2

x = 6.4 / 2 = 3.2

find angle formed by the ladder

cos ∅ = d/l

    ∅ = [tex]cos^{-1][/tex] 2/6.4 = [tex]cos^{-1}[/tex]0.3125  ≈ 71.79⁰

remember the net torque around is = zero

to calculate the magnitude of forces on the ladder we apply the following formula

[tex]N = \frac{mg(dcosteta)+ Mgxcosteta}{lsinteta}[/tex]

m = 11 kg, M = 69 kg, l = 6.4 , x = 3,  teta( ∅ )= 71.79⁰, g = 9.8

back to equation  N = [tex]\frac{11*9.8(2*cos71.79)+ 69*9.8*3* cos71.79}{6.4sin71.79}[/tex]

N = (67.375 + 633.938) / 2.879

N = 243.596 N ≈ 243.6 N

Answer:

11,700Newton

Explanation:

According to Newton's second law, Force = mass × acceleration

Given mass = 65kg.

Acceleration if the car can be gotten using one of the equation of motion as shown.

v² = u²+2as

v is the final velocity = 18m/s

u is the initial velocity = 0m/s

a is the acceleration

s is the distance travelled = 0.9m

On substitution;

18² = 0²+2a(0.9)

18² = 1.8a

a = 324/1.8

a = 180m/²

Net force acting on the body = 65×180

Net force acting on the body = 11,700Newton

Answer:

Kinetic Energy of the disk = 252 J

Explanation:

weight of disk = 810 N

radius = 1.56 m

applied force = 49 N

time = 2.95 s

kinetic energy of disk = ?

first, we find the mass of the disk

mass of disk = weight/acceleration due to gravity(9.81 m/s^2) = 810/9.81 m/s^2

mass of disk = 82.57 kg

torque on the disk = force x radius = 49 x 1.56 = 76.44 N-m

moment of inertia I = m[tex]r^{2}[/tex] = 82.57 x [tex]1.56^{2}[/tex] = 200.9 kg-[tex]m^{2}[/tex]

recall that

Torque T = Iα

where α = angular acceleration

76.44 = 200.9α

α = 76.44/200.9 = 0.38 m/s^2

from the equation of angular motion,

ω = ω' + αt

where ω =  final angular speed

ω' = initial angular speed = 0 rad/s since disk starts from rest

t = time = 2.95 s

imputing values into the equation, we have

ω = 0 + (0.38 x 2.95)

ω = 1.12 rad/s

kinetic energy of the disk = I[tex]w^{2}[/tex]

KE = 200.9 x [tex]1.12^{2}[/tex]

Kinetic Energy of the disk = 252 J